Jury Deliberations: probability

[Hadarmeiri]

If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men? (this question is from MGMAT)

I read the explanation and I understand the overall approach but have a question with, why when you are first looking for the total number of juries that could be randomly selected from the jury pool, you also divide by 3!...i understand teh 15!/12! but where did the 3 come from?

15!/12!3! = 455

This 3! also appears in the next step when we are looking for the number of ways we could select 7 men from a pool of 10 men:

10!/7!3!= 120


WHERE IS THIS 3! COMING FROM?

[hadarmeiri]

sorry! I realize that i really dont know how you get 15!/12! even without that 3!


would you do 15x14x13x12x.... for 12 spaces?

:(

[StaceyKoprince]

Please don't forget to post the full text of the question, including answer choices.

Since this is a probability question, we want to find: (# of desired options) / (total # of options)

The total number of options (the denominator) is that 15!/(12!3!) part. We have 15 people total from whom to choose. We will choose 12 and we won't choose 3. The explanation given is from our anagram grid method in our Word Translations strategy guide. We don't actually use the "official math formula" here, though you certainly can if you prefer.

I'm not sure whether you have our book so don't know whether you know the anagram grid method. It's kind of difficult to show on the formatting-limited forums, unfortunately. The basic idea is that your numerator is your total possible group (15 people in this case) and the denominator is the factorial of the # you do choose (12 in this case) times the factorial of the # you don't choose (3 in this case) - so, 12! and 3! respectively. That gives you 15!/(12!*3!) for the overall denominator of our probability problem, above.

[shaan17]

Is the correct answer of this question 86/455 ? Is the numerator the Prob of having 8 men + 9 men + 10 men ?

[RonPurewal]

Is the correct answer of this question 86/455 ? Is the numerator the Prob of having 8 men + 9 men + 10 men ?

yeah, you'd have to do p(8 men, or 9 men, or 10 men)

the # of ways to pick 8 men (and 4 women) is
(10! / 8!2!) x (5! / 4!1!) -- don't forget that you have to pick the women, too
= 45 x 5 = 225

the # of ways to pick 9 men (and 3 women) is
(10! / 9!1!) x (5! / 3!2!)
= 10 x 10 = 100

the # of ways to pick 10 men (and 2 women) is
(10! / 10!0!) x (5! / 2!3!)
= 1 x 10 = 10

...so the numerator should be 335.

note that 0! is defined as 1. (if you don't like that, you could also figure out that the # of ways to pick 10 men out of 10 is only one, just by thinking about it)

[ashutosh_cabm]

Although i agree to the presented solution, i have one doubt in the solutions to the problem.

As stated in the solution:
Probability of (8m, 4w) = (10C8 * 5C4)/15C12

Why can't be calculated it as follows:
Probability of (8m, 4w) = (8/10)*(4/5)

Both give different numbers. We calculate probability by second method in many cases.

Please clarify why is second method wrong..??

[tim]

I will clarify why the second method is wrong when you write out an English sentence explaining the meaning of each term and each operation in your calculation. Of course, once you have completed this exercise you will probably see what was wrong with your approach on your own. This is a very useful exercise anytime you want to know whether your answer is correct or why your answer is incorrect in a probability problem. Remember every term you use and every step you take must make sense or you have probably done it wrong..